3.918 \(\int \frac{x^2}{1-x^2+x^4} \, dx\)

Optimal. Leaf size=74 \[ \frac{\log \left (x^2-\sqrt{3} x+1\right )}{4 \sqrt{3}}-\frac{\log \left (x^2+\sqrt{3} x+1\right )}{4 \sqrt{3}}-\frac{1}{2} \tan ^{-1}\left (\sqrt{3}-2 x\right )+\frac{1}{2} \tan ^{-1}\left (2 x+\sqrt{3}\right ) \]

[Out]

-ArcTan[Sqrt[3] - 2*x]/2 + ArcTan[Sqrt[3] + 2*x]/2 + Log[1 - Sqrt[3]*x + x^2]/(4*Sqrt[3]) - Log[1 + Sqrt[3]*x
+ x^2]/(4*Sqrt[3])

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Rubi [A]  time = 0.0502575, antiderivative size = 74, normalized size of antiderivative = 1., number of steps used = 9, number of rules used = 6, integrand size = 16, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.375, Rules used = {1127, 1161, 618, 204, 1164, 628} \[ \frac{\log \left (x^2-\sqrt{3} x+1\right )}{4 \sqrt{3}}-\frac{\log \left (x^2+\sqrt{3} x+1\right )}{4 \sqrt{3}}-\frac{1}{2} \tan ^{-1}\left (\sqrt{3}-2 x\right )+\frac{1}{2} \tan ^{-1}\left (2 x+\sqrt{3}\right ) \]

Antiderivative was successfully verified.

[In]

Int[x^2/(1 - x^2 + x^4),x]

[Out]

-ArcTan[Sqrt[3] - 2*x]/2 + ArcTan[Sqrt[3] + 2*x]/2 + Log[1 - Sqrt[3]*x + x^2]/(4*Sqrt[3]) - Log[1 + Sqrt[3]*x
+ x^2]/(4*Sqrt[3])

Rule 1127

Int[(x_)^2/((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[a/c, 2]}, Dist[1/2, Int[(q + x^2)/(
a + b*x^2 + c*x^4), x], x] - Dist[1/2, Int[(q - x^2)/(a + b*x^2 + c*x^4), x], x]] /; FreeQ[{a, b, c}, x] && Lt
Q[b^2 - 4*a*c, 0] && PosQ[a*c]

Rule 1161

Int[((d_) + (e_.)*(x_)^2)/((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[(2*d)/e - b/c, 2]},
Dist[e/(2*c), Int[1/Simp[d/e + q*x + x^2, x], x], x] + Dist[e/(2*c), Int[1/Simp[d/e - q*x + x^2, x], x], x]] /
; FreeQ[{a, b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && EqQ[c*d^2 - a*e^2, 0] && (GtQ[(2*d)/e - b/c, 0] || ( !Lt
Q[(2*d)/e - b/c, 0] && EqQ[d - e*Rt[a/c, 2], 0]))

Rule 618

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> Dist[-2, Subst[Int[1/Simp[b^2 - 4*a*c - x^2, x], x]
, x, b + 2*c*x], x] /; FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 204

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> -Simp[ArcTan[(Rt[-b, 2]*x)/Rt[-a, 2]]/(Rt[-a, 2]*Rt[-b, 2]), x] /
; FreeQ[{a, b}, x] && PosQ[a/b] && (LtQ[a, 0] || LtQ[b, 0])

Rule 1164

Int[((d_) + (e_.)*(x_)^2)/((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[(-2*d)/e - b/c, 2]},
Dist[e/(2*c*q), Int[(q - 2*x)/Simp[d/e + q*x - x^2, x], x], x] + Dist[e/(2*c*q), Int[(q + 2*x)/Simp[d/e - q*x
- x^2, x], x], x]] /; FreeQ[{a, b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && EqQ[c*d^2 - a*e^2, 0] &&  !GtQ[b^2
- 4*a*c, 0]

Rule 628

Int[((d_) + (e_.)*(x_))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Simp[(d*Log[RemoveContent[a + b*x +
c*x^2, x]])/b, x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[2*c*d - b*e, 0]

Rubi steps

\begin{align*} \int \frac{x^2}{1-x^2+x^4} \, dx &=-\left (\frac{1}{2} \int \frac{1-x^2}{1-x^2+x^4} \, dx\right )+\frac{1}{2} \int \frac{1+x^2}{1-x^2+x^4} \, dx\\ &=\frac{1}{4} \int \frac{1}{1-\sqrt{3} x+x^2} \, dx+\frac{1}{4} \int \frac{1}{1+\sqrt{3} x+x^2} \, dx+\frac{\int \frac{\sqrt{3}+2 x}{-1-\sqrt{3} x-x^2} \, dx}{4 \sqrt{3}}+\frac{\int \frac{\sqrt{3}-2 x}{-1+\sqrt{3} x-x^2} \, dx}{4 \sqrt{3}}\\ &=\frac{\log \left (1-\sqrt{3} x+x^2\right )}{4 \sqrt{3}}-\frac{\log \left (1+\sqrt{3} x+x^2\right )}{4 \sqrt{3}}-\frac{1}{2} \operatorname{Subst}\left (\int \frac{1}{-1-x^2} \, dx,x,-\sqrt{3}+2 x\right )-\frac{1}{2} \operatorname{Subst}\left (\int \frac{1}{-1-x^2} \, dx,x,\sqrt{3}+2 x\right )\\ &=-\frac{1}{2} \tan ^{-1}\left (\sqrt{3}-2 x\right )+\frac{1}{2} \tan ^{-1}\left (\sqrt{3}+2 x\right )+\frac{\log \left (1-\sqrt{3} x+x^2\right )}{4 \sqrt{3}}-\frac{\log \left (1+\sqrt{3} x+x^2\right )}{4 \sqrt{3}}\\ \end{align*}

Mathematica [C]  time = 0.141765, size = 94, normalized size = 1.27 \[ \frac{\sqrt{-1-i \sqrt{3}} \left (\sqrt{3}+i\right ) \tan ^{-1}\left (\frac{1}{2} \left (1-i \sqrt{3}\right ) x\right )+\sqrt{-1+i \sqrt{3}} \left (\sqrt{3}-i\right ) \tan ^{-1}\left (\frac{1}{2} \left (1+i \sqrt{3}\right ) x\right )}{2 \sqrt{6}} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[x^2/(1 - x^2 + x^4),x]

[Out]

(Sqrt[-1 - I*Sqrt[3]]*(I + Sqrt[3])*ArcTan[((1 - I*Sqrt[3])*x)/2] + Sqrt[-1 + I*Sqrt[3]]*(-I + Sqrt[3])*ArcTan
[((1 + I*Sqrt[3])*x)/2])/(2*Sqrt[6])

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Maple [A]  time = 0.053, size = 57, normalized size = 0.8 \begin{align*}{\frac{\arctan \left ( 2\,x-\sqrt{3} \right ) }{2}}+{\frac{\arctan \left ( 2\,x+\sqrt{3} \right ) }{2}}+{\frac{\ln \left ( 1+{x}^{2}-x\sqrt{3} \right ) \sqrt{3}}{12}}-{\frac{\ln \left ( 1+{x}^{2}+x\sqrt{3} \right ) \sqrt{3}}{12}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2/(x^4-x^2+1),x)

[Out]

1/2*arctan(2*x-3^(1/2))+1/2*arctan(2*x+3^(1/2))+1/12*ln(1+x^2-x*3^(1/2))*3^(1/2)-1/12*ln(1+x^2+x*3^(1/2))*3^(1
/2)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x^{2}}{x^{4} - x^{2} + 1}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2/(x^4-x^2+1),x, algorithm="maxima")

[Out]

integrate(x^2/(x^4 - x^2 + 1), x)

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Fricas [B]  time = 1.57218, size = 529, normalized size = 7.15 \begin{align*} -\frac{1}{6} \, \sqrt{6} \sqrt{3} \sqrt{2} \arctan \left (-\frac{1}{3} \, \sqrt{6} \sqrt{3} \sqrt{2} x + \frac{1}{3} \, \sqrt{6} \sqrt{3} \sqrt{\sqrt{6} \sqrt{2} x + 2 \, x^{2} + 2} - \sqrt{3}\right ) - \frac{1}{6} \, \sqrt{6} \sqrt{3} \sqrt{2} \arctan \left (-\frac{1}{3} \, \sqrt{6} \sqrt{3} \sqrt{2} x + \frac{1}{3} \, \sqrt{6} \sqrt{3} \sqrt{-\sqrt{6} \sqrt{2} x + 2 \, x^{2} + 2} + \sqrt{3}\right ) - \frac{1}{24} \, \sqrt{6} \sqrt{2} \log \left (\sqrt{6} \sqrt{2} x + 2 \, x^{2} + 2\right ) + \frac{1}{24} \, \sqrt{6} \sqrt{2} \log \left (-\sqrt{6} \sqrt{2} x + 2 \, x^{2} + 2\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2/(x^4-x^2+1),x, algorithm="fricas")

[Out]

-1/6*sqrt(6)*sqrt(3)*sqrt(2)*arctan(-1/3*sqrt(6)*sqrt(3)*sqrt(2)*x + 1/3*sqrt(6)*sqrt(3)*sqrt(sqrt(6)*sqrt(2)*
x + 2*x^2 + 2) - sqrt(3)) - 1/6*sqrt(6)*sqrt(3)*sqrt(2)*arctan(-1/3*sqrt(6)*sqrt(3)*sqrt(2)*x + 1/3*sqrt(6)*sq
rt(3)*sqrt(-sqrt(6)*sqrt(2)*x + 2*x^2 + 2) + sqrt(3)) - 1/24*sqrt(6)*sqrt(2)*log(sqrt(6)*sqrt(2)*x + 2*x^2 + 2
) + 1/24*sqrt(6)*sqrt(2)*log(-sqrt(6)*sqrt(2)*x + 2*x^2 + 2)

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Sympy [A]  time = 0.168825, size = 63, normalized size = 0.85 \begin{align*} \frac{\sqrt{3} \log{\left (x^{2} - \sqrt{3} x + 1 \right )}}{12} - \frac{\sqrt{3} \log{\left (x^{2} + \sqrt{3} x + 1 \right )}}{12} + \frac{\operatorname{atan}{\left (2 x - \sqrt{3} \right )}}{2} + \frac{\operatorname{atan}{\left (2 x + \sqrt{3} \right )}}{2} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**2/(x**4-x**2+1),x)

[Out]

sqrt(3)*log(x**2 - sqrt(3)*x + 1)/12 - sqrt(3)*log(x**2 + sqrt(3)*x + 1)/12 + atan(2*x - sqrt(3))/2 + atan(2*x
+ sqrt(3))/2

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x^{2}}{x^{4} - x^{2} + 1}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2/(x^4-x^2+1),x, algorithm="giac")

[Out]

integrate(x^2/(x^4 - x^2 + 1), x)